Question

Chứng tỏ: 1<1\5+1\6+...+1\16+1\17<2

Answer 1

Xét \(\frac{1}{5}+\frac{1}{6}>\frac{1}{17}.6=\frac{6}{17}\)

và \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+...+\frac{1}{17}>\frac{1}{17}+\frac{1}{17}+\frac{1}{17}+...+\frac{1}{17}=\frac{1}{17}.11=\frac{11}{17}\)

Do đó \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}>\frac{6}{17}+\frac{11}{17}=\frac{17}{17}=1\) (1)

Lại có \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}\right)+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{17}\right)\)

\(< \left(\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\right)+\left(\frac{1}{17}+\frac{1}{17}+...+\frac{1}{17}\right)=\frac{1}{10}.6+\frac{1}{17}.7=1\frac{1}{85}< 2\) (2)

Từ (1) và (2) suy ra điều phải chứng minh